Problem: An acute isosceles triangle, $ABC$, is inscribed in a circle.  Through $B$ and $C$, tangents to the circle are drawn, meeting at point $D$.  If $\angle ABC = \angle ACB = 2 \angle D$ and $\angle BAC = k \pi$ in radians, then find $k$.

[asy]
import graph;

unitsize(2 cm);

pair O, A, B, C, D;

O = (0,0);
A = dir(90);
B = dir(-30);
C = dir(210);
D = extension(B, B + rotate(90)*(B), C, C + rotate(90)*(C));

draw(Circle(O,1));
draw(A--B--C--cycle);
draw(B--D--C);

label("$A$", A, N);
label("$B$", B, SE);
label("$C$", C, SW);
label("$D$", D, S);
[/asy]
Solution: Let $x = \angle BAC$.  Angles $\angle BAC$, $\angle BCD$, and $\angle CBD$ all intercept the same circular arc, minor arc $BC$ with measure $2 \angle BAC = 2x$.  Then $\angle BCD = \angle CBD = x$, so $\angle D = \pi - 2x$.

Since $\angle ABC = \angle ACB$, $\angle ABC = (\pi - x)/2$.  Then from the equation $\angle ABC = 2 \angle D$, \[\frac{\pi - x}{2} = 2 (\pi - 2x).\]Solving for $x$, we find $x = 3 \pi/7$, so $k = \boxed{3/7}$.